If a compact set's boundary is the finite union of smooth Jordan curves, is it a domain?

Question

Let $V \subset \mathbb{R}^2$ be a compact set such that it's boundary is a finite union of piecewise smooth Jordan curves.

For example:

Does this imply that $V \setminus \partial V$ is a domain?

(If yes, does this also apply to higher finite dimensions, namely $\mathbb{R}^n$?)

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EDIT: Thanks to @UmbertoP it is clear that the answer is no, and a simple counter-example is the union of two disjoint closed disks.

Now, if I add the following conditions:

• The jordan curves mentioned are pairwise disjoint (Thanks @LeeMosher)
• One of the jordan curves is the boundary of a set that contains all other jordan curves

Do I have a domain now?

Answer 1

Even if one of the Jordan curves fully encloses all the others, you get counterexamples: Insert a closed disk in one the holes.

Let us first consider a non-connected $V$. If it has two or more components having an interior point, then $V \backslash \partial V$ cannot be connected. Hence $V$ must be required to have at most one component with an interior point. In this case all other components must be Jordan curves.

This shows that we may focus to connected $V$ having an interior point. In this case $V \backslash \partial V$ is clearly connected because $V$ is a $2$-manifold with boundary (in general not a smooth manifold since the Jordan curves are only piecewise smooth). Its interior is connected. See Is the interior of a smooth manifold with boundary connected? (the proof given in the first answer is valid also for topological manifolds).